# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:21:01 AM

Title: Problem 3 (main sitting)
Post by: Victor Ivrii on November 19, 2019, 04:21:01 AM
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} 1 &1\\ -2 &4\end{pmatrix}\mathbf{x}$$
classify fixed point $(0,0)$ and sketch trajectories.

(b) Find the general solution
$$\mathbf{x}'=\begin{pmatrix} 1 &1\\ -2 &4\end{pmatrix}\mathbf{x}+ \begin{pmatrix} \dfrac{e^{4t }}{e^{2t}+1} \\ 0\end{pmatrix}.$$
Title: Re: Problem 3 (main sitting)
Post by: Yiheng Bian on November 19, 2019, 04:29:59 AM
(a):
We solve homo firstly:
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} 1-\lambda & 1 \\ -2 & 4-\lambda \end{vmatrix}=-5\lambda+{\lambda}^2+6=0\\ (\lambda-2)(\lambda-3)=0\\ \lambda_1=2,\lambda_2=3$$
Then:
$$(A-{\lambda}I)x=0$$
$$When \lambda=2$$
$$\begin{pmatrix} -1 & 1 \\ -2 & 2 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
RREF:
$$\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
$$\text{Let x_2=t, so }x_1=x_2=t\\ t*\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad$$
$$When \lambda=3$$
$$\begin{pmatrix} -2 & 1 \\ -2 & 1 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
RREF:
$$\begin{pmatrix} 2 & -1 \\ 0 & 0 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
$$\text{Let x_2=t, so }x_1=0.5t, x_2=t\\ t*\begin{pmatrix} 1 \\ 2 \end{pmatrix} \quad$$
So, the general solution is :
$$y= c_1e^{2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad +c_2e^{3t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} \quad$$

(b):
$$\phi = \begin{pmatrix} e^{2t} & e^{3t} \\ e^{2t} & 2e^{3t} \end{pmatrix} \quad$$
$$\phi * u' = g(t)$$
$$\begin{pmatrix} e^{2t} & e^{3t} \\ e^{2t} & 2e^{3t} \end{pmatrix} \quad *{\begin{pmatrix} u_1' \\ u_2' \end{pmatrix} \quad}=\begin{pmatrix} \frac{e^{4t}}{e^{2t} + 1 } \\ 0 \end{pmatrix} \quad$$
Simplify we can get:
$$u_1'=-\frac{-e^t}{e^{2t} + 1}\\ u_2'=\frac{2e^{2t}}{e^{2t}+1}$$
Therefore:
$$u_1=ln(e^{2t}+1)+c_1\\ u_2=-arctane^t+c_2$$
Finally:
$$x=\phi * u=(ln(e^{2t}+1)+c_1)*{\begin{pmatrix} e^{2t} \\ e^{2t} \end{pmatrix} \quad} +(-arctane^t+c_2)*\begin{pmatrix} e^{3t} \\ 2e^{3t} \end{pmatrix} \quad$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
Title: Re: Problem 3 (main sitting)
Post by: xuanzhong on November 19, 2019, 05:19:08 AM
Here's the solution including sketching.
Title: Re: Problem 3 (main sitting)
Post by: Aparna on November 19, 2019, 08:45:00 AM
Computer-generated sketch:
Title: Re: Problem 3 (main sitting)
Post by: Sifan Shao on November 19, 2019, 09:23:37 AM
antiderivative of u1': let e^t = u, so e^2t+1 = u^2+1, du = e^t dt, so the formula becomes integral( - 1/(u^2+1)du), which is -arctan(u) -> -arctan(e^t)
antiderivative of u2': let e^2t = u, so the formula becomes integral(2u/(u+1)), which is by chain rule: ln(u+1) -> ln(e^2t+1)
Title: Re: Problem 3 (main sitting)
Post by: Mingdi Xie on November 19, 2019, 09:41:06 PM
This is my solution
Title: Re: Problem 3 (main sitting)
Post by: Mingdi Xie on November 19, 2019, 09:44:19 PM
There is a small typo in Yiheng Bian's solution, $u_1'$ and $u_2'$ should be alternate.
Title: Re: Problem 3 (main sitting)
Post by: Victor Ivrii on November 24, 2019, 09:41:32 AM
What everybody is missing:
Part of the problem "classify fixed point $(0,0)$".
It is unstable node,